How do you find the derivative of #sin^2(8x)-(pi)x#?

Question

Emilia Aguilar · Accepted Answer

#frac(d)(dx) ((sin(8 x))^(2) - pi x) = 8 sin(16 x) - pi#

We have: #sin^(2)(8 x) - (pi) x#

#= (sin(8 x))^(2) - pi x#

First, let's apply the difference rule:

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = frac(d)(dx)((sin(8 x))^(2)) - frac(d)(dx)(pi x)#

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = frac(d)(dx) ((sin(8 x))^(2)) - pi#

Then, let's use the chain rule.

Let #u = sin(8 x) Rightarrow u' = cos(8 x)#, #v = u^(2) Rightarrow v' = 2 u# and #w = 8 x Rightarrow w' = 8#:

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = u' cdot v' cdot w' - pi#

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = cos(8 x) cdot 2 u cdot 8 - pi#

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = 8 cdot 2 u cos(8 x) - pi#

Now, let's replace #u# with #sin(8 x)#:

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = 8 cdot 2 sin(8 x) cos(8 x) - pi#

Let's apply the double angle formula for #sin(x)#; #sin(2 x) = 2 sin(x) cos(x)#:

#Rightarrow frac(d)(dx) ((sin(8 x))^(2) - pi x) = 8 cdot sin(2 cdot 8 x) - pi#

#therefore frac(d)(dx) ((sin(8 x))^(2) - pi x) = 8 sin(16 x) - pi#

Charles Anctil · Answer

undefined To find the derivative of $ \sin^2(8x) - (\pi)x $, you would apply the chain rule and the power rule.

First, differentiate $ \sin^2(8x) $ using the chain rule. The derivative of $ \sin^2(u) $ with respect to $ u $ is $ 2\sin(u)\cos(u) $. Then, multiply by the derivative of the inner function, $ 8x $, which is $ 8 $.

Next, differentiate $ -(\pi)x $ using the power rule, where the derivative of $ ax $ with respect to $ x $ is $ a $. So, the derivative of $-( \pi)x$ is $ -\pi $.

Combining these derivatives, the derivative of $ \sin^2(8x) - (\pi)x $ is:

$$ 2\sin(8x)\cos(8x) \cdot 8 - \pi $$

Which simplifies to:

$$ 16\sin(8x)\cos(8x) - \pi $$